What does the mean value theorem tell you about such functions?
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dlsMay 23 '12 at 6:10

@AmihaiZivan: Your example doesn't work since $\frac{1}{1+x^{2}}\leq 1$ for all $x\in\mathbb{R}$ (instead of $\geq 1$!). As Arturo Magidin demonstrated below, a function with such properties does not exist.
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Thomas E.May 23 '12 at 8:13

2 Answers
2

Pick an $a\gt 0$. Then by the Mean Value Theorem, there exists a point $r$, $a\lt r\lt a+M$, such that $f'(r)=\frac{1}{M}(f(a+M)-f(a))$. That means that
$$f(a+M) - f(a) = Mf'(r) \geq Mc$$
hence for every $M\gt 0$ we have
$$f(a+M) \geq f(a)+Mc.$$